Question 442014
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
Hi
Note: vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y=1/2(x+3)^2-1  |a = .5 > 0  Parabola opens upward.
Vertex(-3,-1) Line of symmetry is x = -3
 y = .5*9-1 when x = 0  y-intercept is (0,7/2)
 0 = .5(x+3)^2 - 1
 2 = (x+3)^2
 ±{{{sqrt(2)}}}-3 = x when y = 0  x-intercepts are (-4.4142,0) and (-1.5858, 0)
{{{drawing(300,300,   -6, 6, -6, 6,  blue(line(-3,6,-3,-6))  , grid(1),
circle(-3, -1,0.3),
circle(-4.4142, 0,0.3),
circle(-1.5858, 0,0.3),
circle(0, 3.5,0.3),

graph( 300, 300, -6, 6, -6, 6,0,.5(x+3)^2-1))}}}



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